If you 3.7 grams of BeCl₂ and 20 ml of a 3M solution of LiH and you make 0.22 grams of BeH₂ and .99 grams of LiCl. e. What is your theoretical yield of BeH₂ in grams? f. What is your theoretical yield of LiCl in moles? g. What is your theoretical yield of LiCl in grams? h. What are your percent yields?
If you 3.7 grams of BeCl₂ and 20 ml of a 3M solution of LiH and you make 0.22 grams of BeH₂ and .99 grams of LiCl. e. What is your theoretical yield of BeH₂ in grams? f. What is your theoretical yield of LiCl in moles? g. What is your theoretical yield of LiCl in grams? h. What are your percent yields?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Chemistry Problem: Reaction Yield Calculations**
Given:
- 3.7 grams of BeCl₂
- 20 ml of a 3M solution of LiH
- Products obtained: 0.22 grams of BeH₂ and 0.99 grams of LiCl
**Questions:**
e. What is your theoretical yield of BeH₂ in grams?
f. What is your theoretical yield of LiCl in moles?
g. What is your theoretical yield of LiCl in grams?
h. What are your percent yields?
---
**Explanation and Steps:**
1. **Molar Masses:**
- BeCl₂ (Beryllium Chloride): Beryllium (Be) = 9.012 g/mol; Chlorine (Cl) = 35.453 g/mol
\[
\text{Molar Mass of BeCl₂} = 9.012 + 2 \times 35.453 = 79.918 \, \text{g/mol}
\]
- LiH (Lithium Hydride): Lithium (Li) = 6.941 g/mol; Hydrogen (H) = 1.008 g/mol
\[
\text{Molar Mass of LiH} = 6.941 + 1.008 = 7.949 \, \text{g/mol}
\]
- BeH₂ (Beryllium Hydride): Beryllium (Be) = 9.012 g/mol; Hydrogen (H) = 1.008 g/mol
\[
\text{Molar Mass of BeH₂} = 9.012 + 2 \times 1.008 = 11.028 \, \text{g/mol}
\]
- LiCl (Lithium Chloride): Lithium (Li) = 6.941 g/mol; Chlorine (Cl) = 35.453 g/mol
\[
\text{Molar Mass of LiCl} = 6.941 + 35.453 = 42.394 \, \text{g/mol}
\]
2. **Balanced Chemical Equation:**
\[
\text{BeCl₂} + 2\text{LiH} \rightarrow \text{BeH₂} + 2\text{LiCl}
\]
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Transcribed Image Text:**Chemistry Problem: Reaction Yield Calculations**
Given:
- 3.7 grams of BeCl₂
- 20 ml of a 3M solution of LiH
- Products obtained: 0.22 grams of BeH₂ and 0.99 grams of LiCl
**Questions:**
e. What is your theoretical yield of BeH₂ in grams?
f. What is your theoretical yield of LiCl in moles?
g. What is your theoretical yield of LiCl in grams?
h. What are your percent yields?
---
**Explanation and Steps:**
1. **Molar Masses:**
- BeCl₂ (Beryllium Chloride): Beryllium (Be) = 9.012 g/mol; Chlorine (Cl) = 35.453 g/mol
\[
\text{Molar Mass of BeCl₂} = 9.012 + 2 \times 35.453 = 79.918 \, \text{g/mol}
\]
- LiH (Lithium Hydride): Lithium (Li) = 6.941 g/mol; Hydrogen (H) = 1.008 g/mol
\[
\text{Molar Mass of LiH} = 6.941 + 1.008 = 7.949 \, \text{g/mol}
\]
- BeH₂ (Beryllium Hydride): Beryllium (Be) = 9.012 g/mol; Hydrogen (H) = 1.008 g/mol
\[
\text{Molar Mass of BeH₂} = 9.012 + 2 \times 1.008 = 11.028 \, \text{g/mol}
\]
- LiCl (Lithium Chloride): Lithium (Li) = 6.941 g/mol; Chlorine (Cl) = 35.453 g/mol
\[
\text{Molar Mass of LiCl} = 6.941 + 35.453 = 42.394 \, \text{g/mol}
\]
2. **Balanced Chemical Equation:**
\[
\text{BeCl₂} + 2\text{LiH} \rightarrow \text{BeH₂} + 2\text{LiCl}
\]
3.
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